Revision notes, practice numericals, a question bank, assertion and reason questions and answers to the NCERT questions. Tap a group to open it; opening one closes the others.
When light falls on a polished surface such as a mirror, it bounces back into the same medium. This is reflection. Light seems to travel in straight lines, and reflecting surfaces of all kinds obey the laws of reflection.
A spherical mirror is part of a hollow sphere. If the inner surface is the reflecting surface, it is a concave mirror; if the outer surface reflects, it is a convex mirror.
For spherical mirrors of small aperture, the principal focus lies midway between the pole and the centre of curvature.
R = 2f or f = R / 2The image is formed between F and C. It is real, inverted and diminished.
The image is formed behind the mirror. It is virtual, erect and enlarged. This is how a shaving mirror works.
| Object position | Image position | Size | Nature |
|---|---|---|---|
| At infinity | At F | Highly diminished, point-sized | Real and inverted |
| Beyond C | Between F and C | Diminished | Real and inverted |
| At C | At C | Same size | Real and inverted |
| Between C and F | Beyond C | Enlarged | Real and inverted |
| At F | At infinity | Highly enlarged | Real and inverted |
| Between P and F | Behind the mirror | Enlarged | Virtual and erect |
For every position of the object, a convex mirror forms a virtual, erect and diminished image behind the mirror, between P and F. For an object at infinity the image is at F and highly diminished.
Used as rear-view (wing) mirrors in vehicles. They always give an erect, diminished image, and they have a much wider field of view than a plane mirror, so the driver can see traffic far behind.
So the object distance u is always negative. For a concave mirror f is negative; for a convex mirror f is positive.
The relation between object distance u, image distance v and focal length f is
1/v + 1/u = 1/fMagnification m is the ratio of the height of the image (h') to the height of the object (h). It is also related to distances:
m = h'/h = −v/uWhen light travels obliquely from one transparent medium to another, its direction changes at the boundary. This is refraction. It happens because the speed of light is different in different media.
In a rectangular glass slab, refraction takes place at both the air-glass and glass-air faces. The emergent ray is parallel to the incident ray but is laterally shifted.
The refractive index of a medium (with respect to vacuum) is the ratio of the speed of light in vacuum to its speed in that medium.
n = c / vLight travels at 3 × 10⁸ m/s in vacuum. A medium with a greater refractive index is optically denser and light travels more slowly in it. For example, n for water is 1.33, glass about 1.5 and diamond 2.42.
The image is real, inverted and diminished, between F₂ and 2F₂.
The image is virtual, erect and enlarged, on the same side as the object. A convex lens used like this is a magnifying glass.
| Object position | Image position | Size | Nature |
|---|---|---|---|
| At infinity | At F₂ | Highly diminished, point-sized | Real and inverted |
| Beyond 2F₁ | Between F₂ and 2F₂ | Diminished | Real and inverted |
| At 2F₁ | At 2F₂ | Same size | Real and inverted |
| Between F₁ and 2F₁ | Beyond 2F₂ | Enlarged | Real and inverted |
| At F₁ | At infinity | Highly enlarged | Real and inverted |
| Between O and F₁ | Same side as object | Enlarged | Virtual and erect |
A concave lens always forms a virtual, erect and diminished image on the same side as the object. For an object at infinity the image is at F₁, highly diminished; for an object between infinity and O the image is between F₁ and O.
The same sign convention is used, with distances measured from the optical centre.
1/v − 1/u = 1/fFor a convex lens f is positive; for a concave lens f is negative.
The magnification produced by a lens is the ratio of the image height to the object height.
m = h'/h = v/uA positive m means a virtual, erect image; a negative m means a real, inverted image.
The power of a lens is the reciprocal of its focal length in metres. Its SI unit is the dioptre (D): one dioptre is the power of a lens of focal length 1 metre.
P = 1 / fA convex lens has positive power; a concave lens has negative power. Opticians use power to prescribe lenses.
The power of a combination of lenses placed in contact is the algebraic sum of their powers.
P = P₁ + P₂ + P₃ + …Try each problem first, then open the solution to check your working.
The radius of curvature of a spherical mirror is 20 cm. What is its focal length?
Given: R = 20 cm
Use: f = R / 2
f = 20 / 2 = 10 cm
Answer: 10 cm (negative, −10 cm, if the mirror is concave)
Find the focal length of a convex mirror whose radius of curvature is 32 cm.
Given: R = +32 cm (convex mirror)
Use: f = R / 2
f = 32 / 2 = +16 cm
Answer: +16 cm
A concave mirror produces a three times magnified real image of an object placed at 10 cm in front of it. Where is the image located?
Given: u = −10 cm; m = −3 (real image is inverted)
Use: m = −v / u
−3 = −v / (−10)
−3 = v / 10, so v = −30 cm
Answer: 30 cm in front of the mirror
A concave mirror has a focal length of 12 cm. An object is placed 30 cm in front of it. Find the image position and magnification.
Given: f = −12 cm; u = −30 cm
Use: 1/v = 1/f − 1/u
1/v = −1/12 + 1/30 = (−5 + 2)/60 = −3/60
v = −20 cm
m = −v/u = −(−20)/(−30) = −0.67
Answer: v = −20 cm (real, inverted, diminished); m = −0.67
An object is placed at 10 cm from a convex mirror of focal length 15 cm. Find the position and nature of the image.
Given: u = −10 cm; f = +15 cm
Use: 1/v = 1/f − 1/u
1/v = 1/15 + 1/10 = (2 + 3)/30 = 1/6
v = +6 cm
m = −v/u = −6/(−10) = +0.6
Answer: 6 cm behind the mirror; virtual, erect and diminished (m = +0.6)
An object 5.0 cm long is placed 20 cm in front of a convex mirror of radius of curvature 30 cm. Find the position of the image, its nature and size.
Given: h = 5 cm; u = −20 cm; f = R/2 = +15 cm
Use: 1/v = 1/f − 1/u
1/v = 1/15 + 1/20 = (4 + 3)/60 = 7/60
v = 60/7 = +8.57 cm
m = −v/u = −8.57/(−20) = +0.43
h' = m h = 0.43 × 5 = +2.14 cm
Answer: 8.57 cm behind the mirror; virtual, erect, 2.14 cm tall
An object 7.0 cm tall is placed 27 cm in front of a concave mirror of focal length 18 cm. At what distance should a screen be placed to get a sharp image? Find the size and nature of the image.
Given: h = 7 cm; u = −27 cm; f = −18 cm
Use: 1/v = 1/f − 1/u
1/v = −1/18 + 1/27 = (−3 + 2)/54 = −1/54
v = −54 cm
m = −v/u = −(−54)/(−27) = −2
h' = m h = −2 × 7 = −14 cm
Answer: Screen 54 cm in front of the mirror; real, inverted image 14 cm tall
A concave mirror forms a real image twice the size of the object. The object is 15 cm from the mirror. Find the image distance and the focal length.
Given: u = −15 cm; m = −2
Use: m = −v/u and 1/f = 1/v + 1/u
−2 = −v/(−15), so v = −30 cm
1/f = 1/(−30) + 1/(−15) = −3/30 = −1/10
f = −10 cm
Answer: v = −30 cm; f = −10 cm
A shaving mirror gives an erect image three times as large when the face is 10 cm away. Find the focal length.
Given: u = −10 cm; m = +3 (erect, virtual)
Use: m = −v/u and 1/f = 1/v + 1/u
3 = −v/(−10), so v = +30 cm (behind the mirror)
1/f = 1/30 + 1/(−10) = (1 − 3)/30 = −2/30
f = −15 cm
Answer: −15 cm, a concave mirror
A convex mirror forms an image one-quarter the size of the object, which is 60 cm away. Find the focal length.
Given: u = −60 cm; m = +1/4
Use: m = −v/u and 1/f = 1/v + 1/u
1/4 = −v/(−60), so v = +15 cm
1/f = 1/15 + 1/(−60) = (4 − 1)/60 = 3/60
f = +20 cm
Answer: +20 cm
Light enters from air to glass having refractive index 1.50. What is the speed of light in the glass? The speed of light in vacuum is 3 × 10⁸ m/s.
Given: n = 1.50; c = 3 × 10⁸ m/s
Use: n = c / v
v = c / n = 3 × 10⁸ / 1.50 = 2 × 10⁸ m/s
Answer: 2 × 10⁸ m/s
The refractive index of diamond is 2.42. Find the speed of light in diamond.
Given: n = 2.42; c = 3 × 10⁸ m/s
Use: v = c / n
v = 3 × 10⁸ / 2.42 = 1.24 × 10⁸ m/s
Answer: About 1.24 × 10⁸ m/s
The refractive index of water is 1.33. How fast does light travel in water?
Given: n = 1.33
Use: v = c / n
v = 3 × 10⁸ / 1.33 = 2.26 × 10⁸ m/s
Answer: About 2.26 × 10⁸ m/s
The speed of light in a transparent medium is 1.5 × 10⁸ m/s. Find its refractive index.
Given: v = 1.5 × 10⁸ m/s; c = 3 × 10⁸ m/s
Use: n = c / v
n = 3 × 10⁸ / 1.5 × 10⁸ = 2
Answer: n = 2
How long does light take to cross a 1 m thick block of glass of refractive index 1.5?
Given: n = 1.5; thickness = 1 m
Use: v = c / n and t = distance / v
v = 3 × 10⁸ / 1.5 = 2 × 10⁸ m/s
t = 1 / (2 × 10⁸) = 5 × 10⁻⁹ s
Answer: 5 × 10⁻⁹ s (5 nanoseconds)
The refractive index of glass is 1.50 and of water is 1.33 (both with respect to air/vacuum). Find the refractive index of glass with respect to water.
Given: n_glass = 1.50; n_water = 1.33
Use: n(glass w.r.t. water) = n_glass / n_water
n = 1.50 / 1.33 = 1.13
Answer: About 1.13
A ray of light is incident on a glass surface at 45° and the refracted ray makes an angle of 30° with the normal. Find the refractive index of the glass.
Given: i = 45°; r = 30°
Use: n = sin i / sin r
n = sin 45° / sin 30° = 0.707 / 0.5 = 1.41
Answer: About 1.41
A ray enters glass of refractive index 1.5 at an angle of incidence of 30°. Find the angle of refraction.
Given: n = 1.5; i = 30°
Use: sin r = sin i / n
sin r = 0.5 / 1.5 = 0.333
r = about 19.5° (less than i, since the ray bends towards the normal)
Answer: About 19.5°
An object 5 cm tall is held 25 cm away from a converging lens of focal length 10 cm. Find the position, size and nature of the image.
Given: h = 5 cm; u = −25 cm; f = +10 cm
Use: 1/v = 1/f + 1/u
1/v = 1/10 − 1/25 = (5 − 2)/50 = 3/50
v = +16.7 cm
m = v/u = 16.7/(−25) = −0.67
h' = m h = −0.67 × 5 = −3.3 cm
Answer: Image 16.7 cm on the other side; real, inverted, 3.3 cm tall
A convex lens forms a real and inverted image of a needle at 50 cm from it. Where is the needle placed if the image is the same size as the needle? Also find the power of the lens.
Given: v = +50 cm; |m| = 1, so m = −1
Use: m = v/u and 1/f = 1/v − 1/u
−1 = 50/u, so u = −50 cm: the needle is 50 cm in front of the lens
1/f = 1/50 − 1/(−50) = 2/50 = 1/25
f = +25 cm = 0.25 m
P = 1/f = 1/0.25 = +4 D
Answer: Needle at 50 cm; P = +4 D
A concave lens of focal length 15 cm forms an image 10 cm from the lens. How far is the object placed?
Given: f = −15 cm; v = −10 cm (virtual image on the object side)
Use: 1/u = 1/v − 1/f
1/u = −1/10 + 1/15 = (−3 + 2)/30 = −1/30
u = −30 cm
Answer: 30 cm in front of the lens
Find the power of a concave lens of focal length 2 m.
Given: f = −2 m
Use: P = 1/f
P = 1 / (−2) = −0.5 D
Answer: −0.5 D
Find the focal length of a lens of power −2.0 D. What type of lens is it?
Given: P = −2.0 D
Use: f = 1/P
f = 1/(−2.0) = −0.5 m = −50 cm
A negative focal length means a concave (diverging) lens.
Answer: −50 cm; concave lens
A doctor prescribes a corrective lens of power +1.5 D. Find its focal length. Is the lens diverging or converging?
Given: P = +1.5 D
Use: f = 1/P
f = 1/1.5 = 0.67 m = +67 cm
A positive focal length means a convex, converging lens.
Answer: +67 cm; converging
A convex lens has a focal length of 20 cm. What is its power?
Given: f = +20 cm = 0.20 m
Use: P = 1/f
P = 1/0.20 = +5 D
Answer: +5 D
An object is placed 30 cm from a convex lens of focal length 20 cm. Find where the image forms and its magnification.
Given: u = −30 cm; f = +20 cm
Use: 1/v = 1/f + 1/u
1/v = 1/20 − 1/30 = (3 − 2)/60 = 1/60
v = +60 cm
m = v/u = 60/(−30) = −2
Answer: 60 cm on the other side; real, inverted, twice the size
A magnifying glass has a focal length of 10 cm. An object is placed 6 cm from it. Find the image position and magnification.
Given: u = −6 cm; f = +10 cm
Use: 1/v = 1/f + 1/u
1/v = 1/10 − 1/6 = (3 − 5)/30 = −2/30
v = −15 cm
m = v/u = (−15)/(−6) = +2.5
Answer: Image 15 cm on the same side as the object; virtual, erect, 2.5 times larger
An object is 40 cm from a concave lens of focal length 20 cm. Find the image distance and the magnification.
Given: u = −40 cm; f = −20 cm
Use: 1/v = 1/f + 1/u
1/v = −1/20 − 1/40 = (−2 − 1)/40 = −3/40
v = −13.3 cm
m = v/u = (−13.3)/(−40) = +0.33
Answer: 13.3 cm on the object side; virtual, erect, one-third the size
A lens of power +2.0 D is placed in contact with a lens of power −0.5 D. Find the power and focal length of the combination.
Given: P₁ = +2.0 D; P₂ = −0.5 D
Use: P = P₁ + P₂
P = 2.0 + (−0.5) = +1.5 D
f = 1/P = 1/1.5 = 0.67 m = 67 cm
Answer: +1.5 D; f = 67 cm
A convex lens forms a real image 6 cm tall of an object 3 cm tall, at 30 cm from the lens. Find the focal length.
Given: h = 3 cm; h' = −6 cm; v = +30 cm
Use: m = h'/h = v/u and 1/f = 1/v − 1/u
m = −6/3 = −2, so u = v/m = 30/(−2) = −15 cm
1/f = 1/30 − 1/(−15) = (1 + 2)/30 = 1/10
f = +10 cm
Answer: u = −15 cm; f = +10 cm
Pick a level and answer 10 questions with instant feedback. Each level has a bank of 20 questions, and New 10 questions draws a fresh set.
Read the Assertion (A) and the Reason (R), then pick one of the four standard options. You get 5 questions per set from a bank of 20 at each level, and New 5 questions draws a fresh set.
Answers to every in-text and exercise question of the NCERT chapter. Try each one first, then tap Show answer.
Light rays that are parallel to the principal axis of a concave mirror converge at a specific point on its principal axis after reflection from the mirror. This point is called the principal focus of the concave mirror.
f = R / 2 = 20 / 2 = 10 cm (taken as −10 cm if the mirror is concave and +10 cm if convex).
A concave mirror, when the object is placed between its pole and its principal focus.
A convex mirror always gives an erect, though diminished, image. It also has a wider field of view, because it is curved outwards, so the driver can see a much larger area of traffic behind than with a plane mirror.
R = +32 cm. f = R/2 = 32/2 = +16 cm.
Answer: 16 cm (positive)
u = −10 cm; m = −3 (real image, so inverted).
m = −v/u, so −3 = −v/(−10) and v = −30 cm.
Answer: The image is 30 cm in front of the mirror
It bends towards the normal. Water is optically denser than air, so the speed of light decreases on entering water; a ray going from a rarer to a denser medium bends towards the normal.
v = c/n = 3 × 10⁸ / 1.50 = 2 × 10⁸ m/s.
Answer: 2 × 10⁸ m/s
Light travels fastest in the medium with the lowest refractive index. Water (1.33) has a lower index than kerosene (1.44) and turpentine (1.47), so light travels fastest in water.
The speed of light in diamond is 1/2.42 times its speed in vacuum: v = c/2.42 = about 1.24 × 10⁸ m/s. In other words, light travels 2.42 times more slowly in diamond than in vacuum.
One dioptre is the power of a lens whose focal length is 1 metre. 1 D = 1 m⁻¹.
Image is real, inverted and equal in size: m = −1 and v = +50 cm.
m = v/u gives u = v/m = 50/(−1) = −50 cm: the needle is 50 cm in front of the lens.
1/f = 1/v − 1/u = 1/50 + 1/50 = 1/25, so f = +25 cm = 0.25 m.
P = 1/f = +4 D.
Answer: Needle at 50 cm (at 2F); power = +4 D
f = −2 m; P = 1/f = −0.5 D.
Answer: −0.5 D
A lens must be transparent so that light can pass through. Clay is opaque.
Answer: (d) Clay
A concave mirror gives a virtual, erect and enlarged image only when the object is within the focal length.
Answer: (d) Between the pole of the mirror and its principal focus
At 2F₁ the image forms at 2F₂, is real, inverted and the same size.
Answer: (b) At twice the focal length
A negative focal length indicates a concave mirror and a concave lens (with the new Cartesian sign convention).
Answer: (a) both concave
A plane mirror and a convex mirror always form erect images. A concave mirror forms an erect image only close to it.
Answer: (d) either plane or convex
Small letters are best read with a magnifying glass, which is a convex lens, and the shorter the focal length the greater the magnification.
Answer: (c) A convex lens of focal length 5 cm
The object must be placed between the pole and the principal focus, that is, at a distance less than 15 cm (between 0 and 15 cm) from the mirror.
The image is virtual, erect and larger than the object, formed behind the mirror.
(a) Concave mirror: with the bulb at its focus it sends out a powerful parallel beam.
(b) Convex mirror: it gives an erect, diminished image and a wide field of view.
(c) Concave mirror: it focuses the parallel rays of the Sun at its focus, producing high temperature.
Yes, the lens still produces a complete image, but it is dimmer. Light from every point of the object falls on every part of the lens, and the uncovered half still refracts the rays to the same image point; the image brightness falls because fewer rays pass through.
Experiment: form the image of a candle on a screen, then cover half the lens with black paper; the whole image is still seen on the screen but with less brightness.
h = 5 cm; u = −25 cm; f = +10 cm.
1/v = 1/f + 1/u = 1/10 − 1/25 = 3/50, so v = +16.7 cm.
m = v/u = 16.7/(−25) = −0.67.
h' = m h = −3.3 cm.
Answer: Image at 16.7 cm on the other side of the lens; real, inverted, about 3.3 cm tall (diminished)
f = −15 cm; v = −10 cm.
1/u = 1/v − 1/f = −1/10 + 1/15 = −1/30, so u = −30 cm.
Answer: The object is 30 cm from the lens (image virtual, erect and diminished, on the same side)
u = −10 cm; f = +15 cm.
1/v = 1/f − 1/u = 1/15 + 1/10 = 1/6, so v = +6 cm.
m = −v/u = +0.6.
Answer: Image 6 cm behind the mirror; virtual, erect and diminished (m = +0.6)
The plus sign shows that the image is virtual and erect, and the magnitude 1 shows it is the same size as the object.
f = R/2 = +15 cm; u = −20 cm; h = 5 cm.
1/v = 1/15 + 1/20 = 7/60, so v = +8.57 cm.
m = −v/u = +0.43; h' = +2.14 cm.
Answer: Image 8.57 cm behind the mirror; virtual, erect and diminished (2.14 cm tall)
f = −18 cm; u = −27 cm; h = 7 cm.
1/v = −1/18 + 1/27 = −1/54, so v = −54 cm.
m = −v/u = −2; h' = −14 cm.
Answer: Screen at 54 cm in front of the mirror; image is real, inverted, 14 cm tall (enlarged)
f = 1/P = 1/(−2.0) = −0.5 m = −50 cm.
Answer: −50 cm; a concave (diverging) lens
f = 1/P = 1/1.5 = 0.67 m = +67 cm.
Answer: +67 cm; a convex (converging) lens